Order of transformations to inverse trig function

Question

For graphing $y=\cos^{-1}(1-2x)= \cos^{-1} \left[-2\left(x-\dfrac{1}{2} \right) \right]$

I got this correct by finding the coordinates of the key points but when I trying plotting this just by applying transformations to the basic graph of arccos, I don't get it correct?

For this, we need to translate, then dilate and reflect, right?

I get the correct graph by doing the opposite order, but I'm pretty sure this order is incorrect..

Thanks

Answer 1

If one knows what the graph of $y=\cos^{-1}x$ looks like, one could work it as follows. Afterwards, I will show how to proceed when one does not know what the graph of $y=\cos^{-1}x$ looks like.

1. Begin with the graph of $y=\cos^{-1}x$

2. To graph $y=\cos^{-1}(2x)=\cos^{-1}\left(\dfrac{x}{1/2}\right)$ shrink the graph by a factor of $\frac{1}{2}$ horizontally.

3. To graph $y=\cos^{-1}\left(-\dfrac{x}{1/2}\right)$ reflect the previous graph about the $y$-axis.

4. To graph $y=\cos^{-1}\left(-\dfrac{(x-\frac{1}{2})}{1/2}\right)$, shift the previous graph one-half unit to the right.

Even if one does not know what the graph of $y=\cos^{-1}x$ looks like one can still use transformations to figure it out so long as one know that reflecting a graph $f$ about the line $y=x$ produces the graph of the inverse and that for inverse functions the portion of the graph being reflected must satisfy the horizontal line test: no horizontal line can cross the graph twice. For $f(x)=\cos x$ that is the portion of the graph on the interval $[0.\pi]$.

In that case one would proceed as follows:

If $y=\cos^{-1}(1-2x)$ then $0\le x\le1$ and $0\le y\le\pi$. Solving the equation for $x$ gives

$$ x=\frac{1}{2}(1-\cos y) \text{ for }0\le y\le\pi$$

You may instead graph

$$ y=\frac{1}{2}(1-\cos x) \text{ for }0\le x\le\pi$$

and reflect it in the line $y=x$ to produce the graph.

So your transformations should be made on this second graph.

1. Graph $y=\cos x$ on the interval $[0,\pi]$.

2. Reflect the graph in the $x-axis$ to obtain the graph of $y=-\cos x$.

3. Adding $1$ shifts the graph up one unit to give the graph of $y=1-\cos x$.

4. Multiplying by $\frac{1}{2}$ compresses the graph by half vertically to give the graph of $y=\frac{1}{2}(1-\cos x)$

5. Reflect this graph in the line $y=x$ to obtain the graph of $y=\cos^{-1}(1-2x)$.